A toy cannon is anchored to the ceiling. It shoots a ball staight downward at -1.5 m/s . After 0.6 s , what is the displacement of the ball?

Answer 1

Use the kinematic equation: x = v_it + 1/2at² where v_i = -1.5 m/s and a = -9.8 m/s² and t = 0.6 s to get:

x = (-1.5)(0.6) + 1/2(-9.8)(0.6)²

x = -0.9 + -1.764

x = -2.664 m

So 2.664 m down from the ceiling. To do this problem correctly, we should consider the significant figures and round the answer accordingly. Since the one of the givens has only 1 sig fig (t = 0.6 s), the answer should have 1 sig fig as well which would make it x = -3

Answer 2

This problem is an application of the formula for calculating distance when the initial velocity and acceleration are known. The formula is D = V_0*t + 1/2*g*t^2. where D is the distance travelled, V₀ is the initial velocity, g is the gravitational acceleration, and t is the time. As the initial velocity and the acceleration are in the same direction, the negative sign can be ignored.